Ch 3.9: Forced Vibrations

1
Ch 3.9 Forced Vibrations

• We continue the discussion of the last section,
  and now consider the presence of a periodic
  external force

2
Forced Vibrations with Damping

• Consider the equation below for damped motion and
  external forcing funcion F0cos?t.
• The general solution of this equation has the
  form
• where the general solution of the homogeneous
  equation is
• and the particular solution of the
  nonhomogeneous equation is

3
Homogeneous Solution

• The homogeneous solutions u1 and u2 depend on the
  roots r1 and r2 of the characteristic equation
• Since m, ?, and k are are all positive constants,
  it follows that r1 and r2 are either real and
  negative, or complex conjugates with negative
  real part. In the first case,
• while in the second case
• Thus in either case,

4
Transient and Steady-State Solutions

• Thus for the following equation and its general
  solution,
• we have
• Thus uC(t) is called the transient solution.
  Note however that
• is a steady oscillation with same frequency as
  forcing function.
• For this reason, U(t) is called the steady-state
  solution, or forced response.

5
Transient Solution and Initial Conditions

• For the following equation and its general
  solution,
• the transient solution uC(t) enables us to
  satisfy whatever initial conditions might be
  imposed.
• With increasing time, the energy put into system
  by initial displacement and velocity is
  dissipated through damping force. The motion
  then becomes the response U(t) of the system to
  the external force F0cos?t.
• Without damping, the effect of the initial
  conditions would persist for all time.

6
Rewriting Forced Response

• Using trigonometric identities, it can be shown
  that
• can be rewritten as
• It can also be shown that
• where

7
Amplitude Analysis of Forced Response

• The amplitude R of the steady state solution
• depends on the driving frequency ?. For
  low-frequency excitation we have
• where we recall (?0)2 k /m. Note that F0 /k
  is the static displacement of the spring produced
  by force F0.
• For high frequency excitation,

8
Maximum Amplitude of Forced Response

• Thus
• At an intermediate value of ?, the amplitude R
  may have a maximum value. To find this frequency
  ?, differentiate R and set the result equal to
  zero. Solving for ?max, we obtain
• where (?0)2 k /m. Note ?max lt ?0, and ?max
  is close to ?0 for small ?. The maximum value of
  R is

9
Maximum Amplitude for Imaginary ?max

• We have
• and
• where the last expression is an approximation
  for small ?. If
• ? 2 /(mk) gt 2, then ?max is imaginary. In this
  case, Rmax F0 /k, which occurs at ? 0, and R
  is a monotone decreasing function of ?. Recall
  from Section 3.8 that critical damping occurs
  when ? 2 /(mk) 4.

10
Resonance

• From the expression
• we see that Rmax? F0 /(? ?0) for small ?.
• Thus for lightly damped systems, the amplitude R
  of the forced response is large for ? near ?0,
  since ?max ? ?0 for small ?.
• This is true even for relatively small external
  forces, and the smaller the ? the greater the
  effect.
• This phenomena is known as resonance. Resonance
  can be either good or bad, depending on
  circumstances for example, when building bridges
  or designing seismographs.

11
Graphical Analysis of Quantities

• To get a better understanding of the quantities
  we have been examining, we graph the ratios
  R/(F0/k) vs. ?/?0 for several values of ? ? 2
  /(mk), as shown below.
• Note that the peaks tend to get higher as damping
  decreases.
• As damping decreases to zero, the values of
  R/(F0/k) become asymptotic to ? ?0. Also, if ?
  2 /(mk) gt 2, then Rmax F0 /k,
• which occurs at ? 0.

12
Analysis of Phase Angle

• Recall that the phase angle ? given in the forced
  response
• is characterized by the equations
• If ? ? 0, then cos? ? 1, sin? ? 0, and hence ?
  ? 0. Thus the response is nearly in phase with
  the excitation.
• If ? ?0, then cos? 0, sin? 1, and hence ?
  ? ? /2. Thus response lags behind excitation by
  nearly ? /2 radians.
• If ? large, then cos? ? -1, sin? 0, and hence
  ? ? ? . Thus response lags behind excitation by
  nearly ? radians, and hence they are nearly out
  of phase with each other.

13
Example 1 Forced Vibrations with Damping
(1 of 4)

• Consider the initial value problem
• Then ?0 1, F0 3, and ? ? 2 /(mk) 1/64
  0.015625.
• The unforced motion of this system was discussed
  in Ch 3.8, with the graph of the solution given
  below, along with the graph of the ratios
  R/(F0/k) vs. ?/?0 for different values of ?.

14
Example 1 Forced Vibrations with Damping
(2 of 4)

• Recall that ?0 1, F0 3, and ? ? 2 /(mk)
  1/64 0.015625.
• The solution for the low frequency case ? 0.3
  is graphed below, along with the forcing
  function.
• After the transient response is substantially
  damped out, the steady-state response is
  essentially in phase with excitation, and
  response amplitude is larger than static
  displacement.
• Specifically, R ? 3.2939 gt F0/k 3, and ? ?
  0.041185.

15
Example 1 Forced Vibrations with Damping
(3 of 4)

• Recall that ?0 1, F0 3, and ? ? 2 /(mk)
  1/64 0.015625.
• The solution for the resonant case ? 1 is
  graphed below, along with the forcing function.
• The steady-state response amplitude is eight
  times the static displacement, and the response
  lags excitation by ? /2 radians, as predicted.
  Specifically, R 24 gt F0/k 3, and ? ? /2.

16
Example 1 Forced Vibrations with Damping
(4 of 4)

• Recall that ?0 1, F0 3, and ? ? 2 /(mk)
  1/64 0.015625.
• The solution for the relatively high frequency
  case ? 2 is graphed below, along with the
  forcing function.
• The steady-state response is out of phase with
  excitation, and response amplitude is about one
  third the static displacement.
• Specifically, R ? 0.99655 ? F0/k 3, and ? ?
  3.0585 ? ?.

17
Undamped Equation General Solution for the Case
?0 ? ?

• Suppose there is no damping term. Then our
  equation is
• Assuming ?0 ? ?, then the method of undetermined
  coefficients can be use to show that the general
  solution is

18
Undamped Equation Mass Initially at Rest (1 of
3)

• If the mass is initially at rest, then the
  corresponding initial value problem is
• Recall that the general solution to the
  differential equation is
• Using the initial conditions to solve for c1 and
  c2, we obtain
• Hence

19
Undamped Equation Solution to Initial Value
Problem (2 of 3)

• Thus our solution is
• To simplify the solution even further, let A
  (?0 ?)/2 and B (?0 - ?)/2. Then A B ?0t
  and A - B ?t. Using the trigonometric identity
• it follows that
• and hence

20
Undamped Equation Beats (3 of 3)

• Using the results of the previous slide, it
  follows that
• When ?0 - ? ? 0, ?0 ? is much larger than ?0
  - ?, and
• sin(?0 ?)t/2 oscillates more rapidly than
  sin(?0 - ?)t/2.
• Thus motion is a rapid oscillation with frequency
  (?0 ?)/2, but with slowly varying sinusoidal
  amplitude given by
• This phenomena is called a beat.
• Beats occur with two tuning forks of
• nearly equal frequency.

21
Example 2 Undamped Equation,Mass Initially at
Rest (1 of 2)

• Consider the initial value problem
• Then ?0 1, ? 0.8, and F0 0.5, and hence the
  solution is
• The displacement of the springmass system
  oscillates with a
• frequency of 0.9, slightly less than natural
  frequency ?0 1.
• The amplitude variation has a slow frequency of
  0.1 and period of 20?.
• A half-period of 10? corresponds to a single
  cycle of increasing and then decreasing
  amplitude.

22
Example 2 Increased Frequency (2 of 2)

• Recall our initial value problem
• If driving frequency ? is increased to ? 0.9,
  then the slow frequency is halved to 0.05 with
  half-period doubled to 20?.
• The multiplier 2.77778 is increased to 5.2632,
  and the fast frequency only marginally increased,
  to 0.095.

23
Undamped Equation General Solution for the Case
?0 ? (1 of 2)

• Recall our equation for the undamped case
• If forcing frequency equals natural frequency of
  system, i.e., ? ?0 , then nonhomogeneous term
  F0cos?t is a solution of homogeneous equation.
  It can then be shown that
• Thus solution u becomes unbounded as t ? ?.
• Note Model invalid when u getslarge, since we
  assume small oscillations u.

24
Undamped Equation Resonance (2 of 2)

• If forcing frequency equals natural frequency of
  system, i.e., ? ?0 , then our solution is
• Motion u remains bounded if damping present.
  However, response u to input F0cos?t may be
  large if damping is small and ?0 - ? ? 0, in
  which case we have resonance.